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I would like your help to double check my derivations below involving the joint probability distribution of some Gumbel differences.

Consider $K$ i.i.d. random variables $\epsilon_1,...,\epsilon_K$, where $\epsilon_1$ is distributed as a Gumbel with scale $\beta$ and location $\mu$. Hence the pdf of $\epsilon_1$ is $$ f(x)=\frac{1}{\beta}\exp\Big(-\Big(\frac{x-\mu}{\beta}\Big)\Big)\times \exp\Big(-\exp\Big(-\Big(\frac{x-\mu}{\beta}\Big)\Big)\Big) $$ and the cdf of $\epsilon_1$ is $$ F(x)=\exp\Big(-\exp\Big(-\Big(\frac{x-\mu}{\beta}\Big)\Big)\Big) $$

Question: Let $\alpha_1,...\alpha_K$ be some known real numbers. Derive $$ Prob(\epsilon_1+\alpha_1\geq \epsilon_k+\alpha_k \text{ }\forall k\neq 1) $$


My attempt (could you check it?)

Step 1: $$ Prob(\epsilon_1+\alpha_1\geq \epsilon_k+\alpha_k \text{ }\forall k\neq 1|\epsilon_1)=\Pi_{k\neq 1} Prob(\epsilon_k\leq \alpha_1-\alpha_k+\epsilon_1 |\epsilon_1)= $$ $$ \exp\Big(-\exp\Big(-\Big(\frac{\epsilon_1+\alpha_1-\alpha_k-\mu_x}{\beta_x}\Big)\Big)\Big) $$

Step 2: $$ Prob(\epsilon_1+\alpha_1\geq \epsilon_k+\alpha_k \text{ }\forall k\neq 1)=\int_{-\infty}^{\infty}\Pi_{k\neq 1}\exp\Big(-\exp\Big(-\Big(\frac{t+\alpha_1-\alpha_k-\mu_x}{\beta_x}\Big)\Big)\Big) \times \frac{1}{\beta}\exp\Big(-\Big(\frac{t-\mu}{\beta}\Big)\Big)\times \exp\Big(-\exp\Big(-\Big(\frac{t-\mu}{\beta}\Big)\Big)\Big)dt $$

Step 3: Notice that for $k=1$ $$ \exp\Big(-\exp\Big(-\Big(\frac{t+\alpha_1-\alpha_k-\mu}{\beta}\Big)\Big)\Big)=\exp\Big(-\exp\Big(-\Big(\frac{t-\mu}{\beta}\Big)\Big)\Big) $$

Step 4: apply step 3 to step 2 $$ Prob(\epsilon_1+\alpha_1\geq \epsilon_k+\alpha_k \text{ }\forall k\neq 1)=\int_{-\infty}^{\infty}\Pi_{k} \exp\Big(-\exp\Big(-\Big(\frac{t+\alpha_1-\alpha_k-\mu_x}{\beta_x}\Big)\Big)\Big) \times \frac{1}{\beta}\exp\Big(-\Big(\frac{t-\mu}{\beta}\Big)\Big)dt $$ Step 5: Notice that $$ \Pi_{k} \exp\Big(-\exp\Big(-\Big(\frac{t+\alpha_1-\alpha_k-\mu_x}{\beta_x}\Big)\Big)\Big)=\exp\Big(-\exp\Big(-\Big(\frac{t-\mu}{\beta}\Big)\Big) \underbrace{\sum_{k}\exp\Big(-\Big(\frac{\alpha_1-\alpha_k}{\beta}\Big)\Big)}_{\equiv Q}\Big) $$

Step 6: apply step 5 to step 4 $$ Prob(\epsilon_1+\alpha_1\geq \epsilon_k+\alpha_k \text{ }\forall k\neq 1)=\int_{-\infty}^{\infty}\exp\Big(-\exp\Big(-\Big(\frac{t-\mu}{\beta}\Big)\Big)\times Q\Big) \times \frac{1}{\beta}\exp\Big(-\Big(\frac{t-\mu}{\beta}\Big)\Big)dt $$

Step 7: change variables $$ \omega\equiv \exp\Big(-\Big(\frac{t-\mu}{\beta}\Big)\Big) $$ so that $$ Prob(\epsilon_1+\alpha_1\geq \epsilon_k+\alpha_k \text{ }\forall k\neq 1)=\int_{0}^{\infty}\exp(-\omega Q)\times \frac{1}{\beta}\omega \frac{1}{\omega} d\omega=\frac{1}{\beta}\times \frac{1}{Q} $$ Hence $$ Prob(\epsilon_1+\alpha_1\geq \epsilon_k+\alpha_k \text{ }\forall k\neq 1)=\frac{1}{\beta \times \sum_{k}\exp\Big(-\Big(\frac{\alpha_1-\alpha_k}{\beta}\Big)\Big)}=\frac{\exp(\frac{\alpha_1}{\beta})}{\beta \sum_{k}\exp\Big(\frac{\alpha_k}{\beta}\Big)\Big)} $$

I'm wondering why $\mu$ does not play a role. I'm not even convinced by the $\beta$ at the denominator in the final expression.

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